# CHAP 8.pmd

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##### Scene 1 (0s)

[Audio] In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and cars move. Atoms, molecules, planets, stars and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don't we directly perceive the motion of the earth? An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road– side perceives the bus alongwith the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate? Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion. Activity ______________ 8.1 • Discuss whether the walls of your classroom are at rest or in motion. Activity ______________ 8.2 • Have you ever experienced that the train in which you are sitting appears to move while it is at rest? • Discuss and share your experience. Think and Act We sometimes are endangered by the motion of objects around us, especially if that motion is erratic and uncontrolled as observed in a flooded river, a hurricane or a tsunami. On the other hand, controlled motion can be a service to human beings such as in the generation of hydro-electric power. Do you feel the necessity to study the erratic motion of some objects and learn to control them? 8.1 Describing Motion We describe the location of an object by specifying a reference point. Let us understand this by an example. Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin. 8 M M M M MOTION OTION OTION OTION OTION Chapter 2020- 21

In everyday life, we see some objects at rest and others in motion. Birds fly, fish swim, blood flows through veins and arteries, and cars move. Atoms, molecules, planets, stars and galaxies are all in motion. We often perceive an object to be in motion when its position changes with time. However, there are situations where the motion is inferred through indirect evidences. For example, we infer the motion of air by observing the movement of dust and the movement of leaves and branches of trees. What causes the phenomena of sunrise, sunset and changing of seasons? Is it due to the motion of the earth? If it is true, why don’t we directly perceive the motion of the earth? An object may appear to be moving for one person and stationary for some other. For the passengers in a moving bus, the roadside trees appear to be moving backwards. A person standing on the road–side perceives the bus alongwith the passengers as moving. However, a passenger inside the bus sees his fellow passengers to be at rest. What do these observations indicate? Most motions are complex. Some objects may move in a straight line, others may take a circular path. Some may rotate and a few others may vibrate. There may be situations involving a combination of these. In this chapter, we shall first learn to describe the motion of objects along a straight line. We shall also learn to express such motions through simple equations and graphs. Later, we shall discuss ways of describing circular motion. Activity ______________ 8.1 • Discuss whether the walls of your classroom are at rest or in motion. Activity ______________ 8.2 • Have you ever experienced that the train in which you are sitting appears to move while it is at rest? • Discuss and share your experience. Think and Act We sometimes are endangered by the motion of objects around us, especially if that motion is erratic and uncontrolled as observed in a flooded river, a hurricane or a tsunami. On the other hand, controlled motion can be a service to human beings such as in the generation of hydro-electric power. Do you feel the necessity to study the erratic motion of some objects and learn to control them? 8.1 Describing Motion We describe the location of an object by specifying a reference point. Let us understand this by an example. Let us assume that a school in a village is 2 km north of the railway station. We have specified the position of the school with respect to the railway station. In this example, the railway station is the reference point. We could have also chosen other reference points according to our convenience. Therefore, to describe the position of an object we need to specify a reference point called the origin. 8 M M M M MOTION OTION OTION OTION OTION Chapter 2020-21

##### Scene 2 (3m 27s)

[Audio] 8.1. 1 MOTION ALONG A STRAIGHT LINE The simplest type of motion is the motion along a straight line. We shall first learn to describe this by an example. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point ( Fig. 8.1). Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B. displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time. Activity ______________ 8.3 • Take a metre scale and a long rope. • Walk from one corner of a basket-ball court to its oppposite corner along its sides. • Measure the distance covered by you and magnitude of the displacement. • What difference would you notice between the two in this case? Activity ______________ 8.4 • Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km. • Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India. The total path length covered by the object is OA + AC, that is 60 km + 35 km = 95 km. This is the distance covered by the object. To describe distance we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values. The numerical value of a physical quantity is its magnitude. From this example, can you find out the distance of the final position C of the object from the initial position O? This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement. Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in (Fig. 8.1). For motion of the object from O to A, the distance covered is 60 km and the magnitude of displacement is also 60 km. During its motion from O to A and back to B, the distance covered = 60 km + 25 km = 85 km Fig. 8.1: Positions of an object on a straight line path while the magnitude of displacement = 35 km. Thus, the magnitude of displacement (35 km) is not equal to the path length (85 km). Further, we will notice that the magnitude of the displacement for a course of motion may be zero but the corresponding distance covered is not zero. If we consider the object to travel back to O, the final position concides with the initial position, and therefore, the displacement is zero. However, the distance covered in this journey is OA + AO = 60 km + 60 km = 120 km. Thus, two different physical quantities — the distance and the MOTION 99 2020- 21

8.1.1 MOTION ALONG A STRAIGHT LINE

The simplest type of motion is the motion along a straight line. We shall first learn to describe this by an example. Consider the motion of an object moving along a straight path. The object starts its journey from O which is treated as its reference point (Fig. 8.1). Let A, B and C represent the position of the object at different instants. At first, the object moves through C and B and reaches A. Then it moves back along the same path and reaches C through B.

displacement, are used to describe the overall motion of an object and to locate its final position with reference to its initial position at a given time.

Activity ______________ 8.3

• Take a metre scale and a long rope.

• Walk from one corner of a basket-ball court to its oppposite corner along its sides.

• Measure the distance covered by you and magnitude of the displacement.

• What difference would you notice between the two in this case?

Activity ______________ 8.4

• Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km.

• Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of India.

The total path length covered by the object is OA + AC, that is 60 km + 35 km = 95 km. This is the distance covered by the object. To describe distance we need to specify only the numerical value and not the direction of motion. There are certain quantities which are described by specifying only their numerical values. The numerical value of a physical quantity is its magnitude. From this example, can you find out the distance of the final position C of the object from the initial position O? This difference will give you the numerical value of the displacement of the object from O to C through A. The shortest distance measured from the initial to the final position of an object is known as the displacement. Can the magnitude of the displacement be equal to the distance travelled by an object? Consider the example given in (Fig. 8.1). For motion of the object from O to A, the distance covered is 60 km and the magnitude of displacement is also 60 km. During its motion from O to A and back to B, the distance covered = 60 km + 25 km = 85 km

Fig. 8.1: Positions of an object on a straight line path

while the magnitude of displacement = 35 km. Thus, the magnitude of displacement (35 km) is not equal to the path length (85 km). Further, we will notice that the magnitude of the displacement for a course of motion may be zero but the corresponding distance covered is not zero. If we consider the object to travel back to O, the final position concides with the initial position, and therefore, the displacement is zero. However, the distance covered in this journey is OA + AO = 60 km + 60 km = 120 km. Thus, two different physical quantities — the distance and the

MOTION 99

2020-21

8.1.1 A'10'1T0N ALONG A STRAIGHT LINE The simplest tvoe of motion is the motion a line. We sli,dl first learn to d' this by an examv:le, Consu:ler tne mnnon of an nhmect rnnvme a]nne a srrnw.nt pal-h. The oblect starts iLS journey from O w. lich is treated as its reference point (F [g. 8.1). Let A, B and C represent the of The Obipc•.t at (litterent Instants. At first, the ohlect moves rnroupn C anci B and reaches A. T lien it back along Lile same pach and reaches C throusli B. o while the Of = 'lh11S, the 01 Rrn) is not equal LO the paun leuetil (65 kun. Further, we will notice that the of the d, tor a course at may be zero but tne corresponflino covered is not mrO. IT we. consioer the oölent to travel baek to O. the concuues with the arid thereture, the :mernpnt IS zero. However, the •tance coverea In journey OA + AO = + 60 km 120 Krn, Thus, two different Dhvsical the distance and the quantities B 10 20 30 Fig. 8.1: Positions of an "'•iect a straigkUirL Oath The total path lenr*h covered by the r '*ft. This is the covered hy "T describe distance we need to sp numerical value and not the -ecuon of There are certain nuantiti!ts which are descrtned by spes g only tnelr numer.eal vanmes. The pi.vsieal quanücy is its maenitude. FIOL Ch's can you find out Ice if the fi.'l C 0! the or•iect trom u. initial pastnon '[his nil terenoe you rne numerical value of the Aic p. -lent of the 01 s,ect from O to C throuk' A. The shortest disl ance measured the inltial to the or an omet known as tne displacement. Can the rr of the disolacement be equal to he distance travelled bv an 0htect? tne exempl.e given in (k g. 8.1). For motion of the ontect from O to A, the distance covered is 60 km and the mt1Qtitude of is 60 km. its motion trom O to A and bark to B. Monon km disnlacen. e. used to describe the overall muuon Of obiect and to locate its final pus re.crence to its illif.ial pusilion Activity 8.3 Talee a metre scale and a long Walk from one corner Of a basKcL-ba_ll court to its oppposite corner along its the covered hy you anu ruaez utude 01 Lile dr:uacement. What d .erence would you notice be-Eween the two In tnls case? Activity 8.4 are With a car is driven from Bituba_llcsliw-ar to New The rerenee net-ween the berween r innesnwar antl Aew uemi

reaches Then it mnvps bark along the same paLh anå rea_eneS C Ulrougn B. o 60 km 120 km, Thus, two different phl the distance and quantities B 10 15 20 25 30 Fig. 8.1: Positions of an "'•iect a straig1:lirL Oath displacen. u e used to describe the o The total oath rth covered bv the 'Die and to Ionace Its rererence to its irnual po This is the djs cance covered by the Qt. 'To nescrtrxe (Itstance we need to sp my the numerical value and not the o. -ecnnn of There are cert_ain nuanulics are described by only their numerical values. The eal valur o a pnvsuaal quannrvis its maontnvie. Fron rhts exar1101e. can you find OUL Ice i_he Lj0Si C Of I oblect fl-uLL1 t,L. postxÄun O'? This d: u•rence •five you tile numertoal value ot tne "cev »ent of tne nmecT trom O to C throne A- Tne shortest ulNLance rneasnred f. .11 initial to ehc linal posi€ion of an ohtes+ known as the displacement. can the m Ae of the diqnlacernent be equal to he distance travelled by an obiect? Consider the exam»le given in 8.1). For motion of the o' from O to at a -n time, Activity 8. T."ke a metre scale and a long rope W'IIR from one corner Oi court to its opppusLte corner along WI easure the distance covered bv at 01 the u uat d.. .eaence woulu vuU not between the two In this case? Activity 8. Automobiles are fitted with a dev gnaws rhe travelleci Car is driven rrorn Fsnnhanegnwar

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[Audio] SCIENCE 100 uestions 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. 8. 1.2 UNIFORM MOTION AND NONUNIFORM MOTION Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion. Activity ______________ 8.5 • The data regarding the motion of two different objects A and B are given in Table 8.1. • Examine them carefully and state whether the motion of the objects is uniform or non-uniform. Q (a) (b) Fig. 8.2 Table 8.1 Time Distance Distance travelled by travelled by object A in m object B in m 9:30 am 10 12 9: 45 am 20 19 10:00 am 30 23 10: 15 am 40 35 10: 30 am 50 37 10:45 am 60 41 11: 00 am 70 44 8.2 Measuring the Rate of Motion 2020- 21

SCIENCE 100 uestions 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position? 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. 8.1.2 UNIFORM MOTION AND NON- UNIFORM MOTION Consider an object moving along a straight line. Let it travel 5 m in the first second, 5 m more in the next second, 5 m in the third second and 5 m in the fourth second. In this case, the object covers 5 m in each second. As the object covers equal distances in equal intervals of time, it is said to be in uniform motion. The time interval in this motion should be small. In our day-to-day life, we come across motions where objects cover unequal distances in equal intervals of time, for example, when a car is moving on a crowded street or a person is jogging in a park. These are some instances of non-uniform motion. Activity ______________ 8.5 • The data regarding the motion of two different objects A and B are given in Table 8.1. • Examine them carefully and state whether the motion of the objects is uniform or non-uniform. Q (a) (b) Fig. 8.2 Table 8.1 Time Distance Distance travelled by travelled by object A in m object B in m 9:30 am 10 12 9:45 am 20 19 10:00 am 30 23 10:15 am 40 35 10:30 am 50 37 10:45 am 60 41 11:00 am 70 44 8.2 Measuring the Rate of Motion 2020-21

##### Scene 4 (9m 46s)

[Audio] MOTION 101 Look at the situations given in Fig. 8.2. If the bowling speed is 143 km h–1 in Fig. 8.2(a) what does it mean? What do you understand from the signboard in Fig. 8.2(b)? Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s–1 or m/s. The other units of speed include centimetre per second (cm s–1) and kilometre per hour (km h–1). To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is, average speed = Total distance travelled Total time taken If an object travels a distance s in time t then its speed v is, v = s t ( 8.1) Let us understand this by an example. A car travels a distance of 100 km in 2 h. Its average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. Sometimes it might have travelled faster and sometimes slower than this. Example 8.1 An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object? Solution: Total distance travelled by the object = 16 m + 16 m = 32 m Total time taken = 4 s + 2 s = 6 s Average speed = Total distance travelled Total time taken = 32 m 6 s = 5.33 m s– 1 Therefore, the average speed of the object is 5.33 m s–1. 8.2.1 SPEED WITH DIRECTION The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object's speed, direction of motion or both. When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed. In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is, average velocity = initial velocity + final velocity 2 Mathematically, vav = u + v 2 (8.2) where vav is the average velocity, u is the initial velocity and v is the final velocity of the object. Speed and velocity have the same units, that is, m s–1 or m/s. Activity ______________ 8.6 • Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4 km h–1, estimate the distance of the bus stop or school from your house. 2020- 21

MOTION 101

Look at the situations given in Fig. 8.2. If the bowling speed is 143 km h–1 in Fig. 8.2(a) what does it mean? What do you understand from the signboard in Fig. 8.2(b)? Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. The rate at which objects move can be different. Also, different objects can move at the same rate. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed. The SI unit of speed is metre per second. This is represented by the symbol m s–1 or m/s. The other units of speed include centimetre per second (cm s–1) and kilometre per hour (km h–1). To specify the speed of an object, we require only its magnitude. The speed of an object need not be constant. In most cases, objects will be in non-uniform motion. Therefore, we describe the rate of motion of such objects in terms of their average speed. The average speed of an object is obtained by dividing the total distance travelled by the total time taken. That is,

average speed = Total distance travelled

Total time taken

If an object travels a distance s in time t then its speed v is,

v = s

t (8.1)

Let us understand this by an example. A car travels a distance of 100 km in 2 h. Its average speed is 50 km h–1. The car might not have travelled at 50 km h–1 all the time. Sometimes it might have travelled faster and sometimes slower than this.

Example 8.1 An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?

Solution:

Total distance travelled by the object = 16 m + 16 m = 32 m Total time taken = 4 s + 2 s = 6 s

Average speed =

Total distance travelled

Total time taken

=

32 m

6 s = 5.33 m s–1

Therefore, the average speed of the object is 5.33 m s–1.

8.2.1 SPEED WITH DIRECTION

The rate of motion of an object can be more comprehensive if we specify its direction of motion along with its speed. The quantity that specifies both these aspects is called velocity. Velocity is the speed of an object moving in a definite direction. The velocity of an object can be uniform or variable. It can be changed by changing the object’s speed, direction of motion or both. When an object is moving along a straight line at a variable speed, we can express the magnitude of its rate of motion in terms of average velocity. It is calculated in the same way as we calculate average speed. In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time. That is,

average velocity = initial velocity + final velocity

2

Mathematically, vav =

u + v

2

(8.2)

where vav is the average velocity, u is the initial velocity and v is the final velocity of the object. Speed and velocity have the same units, that is, m s–1 or m/s.

Activity ______________ 8.6

• Measure the time it takes you to walk from your house to your bus stop or the school. If you consider that your average walking speed is 4 km h–1, estimate the distance of the bus stop or school from your house.

2020-21

Look at the situations given in Fig. 8.2. If the ) 43 Rrn n-iin 8,21a) what does it mean? W hat do you unaersra_nd from the stenbuard in 8.21b)? Different Obiects may take different amounts of time to cover a given distance. Some of them move tast arid some move slowlv. The rate at which obieets move can be dt: rerent. Also. dttterent 0btects can move at the same rate. One oi the wavs of measuring the rare Of motion of an oblet':t. is to iind Lhe distance travclied bv Che ubiect in unit time. This quantitv is rekerred to as speed. The SI unit of speed is metre per second. This is represented by the symnol m s-l or m,' s. ?ihe orner units of speed centimetre ner second (cm s-l) ano Kliorne•re per hour ikm h-l). To the sveed Oi an obiect, we require only its makunude. The speed of an ol:gject need not be constant. In most cases. omects be in non-unuorm motion. Thererore, we describe the rate of motion of such obiects in terms of their averaee soeed. The averaue speed of an 0b; -ct is oi_ituined by dividivæ the total travelled by the total time taken. "I'hE is If an ohiect travels a distance s in tilde t then Let us understand this bv an e. •un 'le. A car travels a dist*tnce of 100 km in h. Its .01 the ume. Surnetunes it have tria "„iled laster and and thei_ ano':ber 16 m 2 s. What is the speed of the object? Total distence travelled by the object - Monon 32 m Therefore, the average speed of the ouject 82.1 SPEED WITH DIRECTION 'lhe rate of motion of an object can be more conmpreher.sive if we its direction of mocii wilh its speed. 'The qunntity that spemttes bi these aspects IS ca 'lect velocity. VpJ0C.tty rne speed ot an orner n. in a dennjre dlreeuon. Tne An ohier.t can be unnorm or variable. t be champed by char.utne the Obie •t's oecd, direcuon of mo' ion or both. •n obiect is moving a stran*kii. Hn a variable speed, we e -press L •e _ na_nitude of its rate of ir ot average velocity. It is uculated th same way as we calculate In c. ee the velocity of the object is g at a uniform rate, then average oc. is given by the arithmetic mean of 'ait; A1 velocity and final velocity for a given period of tune. That is, 2 Mathernatically, v 2 wyrere v Is the average velocitv. u is the initial velocilv and V is the velocity 01' the oblcet. Speed and velocity have the same units, Activity 8.6 Measure the time it takes vou to walk from your house to your bus stop or the scnool. It vou conslrter tnat your averaue speed Is 4 km h-' estirr_ ite Che distance of the bus stop or scnøoi Irom your house.

to out tne (ll•tance Travelled nmect in unit time. 'I-ms quanrn.v is rererred to as soecd. The SI unit of soced is metre per second. This is represented bv the symbol m s-l or m/s. 'Ihe ot-kier units ot speed ide centimetre per second (cm s-l) and kilometre per hour (xm n-l). uzo spemfy the speea 01 an oDiect, we reauire its maunuude. The sueed of an oi.neet need not be cunslnnt. In fnust ohiccts will be in nun-unnorm motion. "Iloeretore. we describe the rate of motion of suotl ohiects in terms of their avernøe_ s,oeed. The avera2e speeci of an Obi -ct is oi-na_llled bv divruine the total travelled by the total tune taken. The is, If an ohiect travels a distance s in ti.ae t then Let us understand this hv an e. am ne. A car travels a oi 100 km in h. Its not h.hve trasrellpd at 50 In -dl the tnne. Sommtmes It have treziled raster and someumes slower th n this. thei_ lher 16 m in 2 s. What is the speed of the object? 'lhe rate of motion of an oblect can be cornorenensive if we specriv its directi mouun ai011E-i wiLh its Sueed. The ouanLiL Sijecilies i:hese is Ca vel VeiociLy is the speed of an 01 sier n. avin definite direction. The velof"+v an can be uniform or variable. t cap. be Ch. by changing the •t's direct mol.on or both. Wnnn •n ompct is o. a straiøni lin• vananle spee ir tp- ot avcraue velocitv. In c. ee the velocity of the obit ch', g at a unitorm rate, then av is given by the aritbmenc me '-nr" veiomry and final velocity for a period of cime. Inat is, 2 Mathernatically, v s#ere vm, is the velmit.y, u is the ve1001tv med Tne nnaL veL001ty 01 rne o Speea and velocity nave the same Activity Measure the time. it takes you to w

##### Scene 5 (14m 14s)

[Audio] SCIENCE 102 = 50 km 1000m 1h × × h 1km 3600s = 13.9 m s–1 The average speed of the car is 50 km h–1 or 13.9 m s–1. Example 8.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha. Solution: Total distance covered by Usha in 1 min is 180 m. Displacement of Usha in 1 min = 0 m Average speed = Total distance covered Totaltimetaken = 180m 180 m 1 min = × 1min 1min 60s = 3 m s-1 Average velocity = Displacement Totaltimetaken = 0m 60 s = 0 m s–1 The average speed of Usha is 3 m s–1 and her average velocity is 0 m s–1. 8.3 Rate of Change of Velocity During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero. Can we now express the change in velocity of an object? Activity ______________ 8.7 • At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning. • Can you answer why this happens? • Measure this time interval using a digital wrist watch or a stop watch. • Calculate the distance of the nearest point of lightning. ( Speed of sound in air = 346 m s-1.) uestions 1. Distinguish between speed and velocity. 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? 3. What does the odometer of an automobile measure? 4. What does the path of an object look like when it is in uniform motion? 5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 m s–1. Example 8.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1. Solution: Distance covered by the car, s = 2400 km – 2000 km = 400 km Time elapsed, t = 8 h Average speed of the car is, vav = 400 km 8 h = s t = 50 km h–1 Q 2020- 21

SCIENCE 102

= 50

km 1000m 1h

× ×

h 1km 3600s

= 13.9 m s–1

The average speed of the car is 50 km h–1 or 13.9 m s–1.

Example 8.3 Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha. Solution: Total distance covered by Usha in 1 min is 180 m. Displacement of Usha in 1 min = 0 m

Average speed =

Total distance covered

Totaltimetaken

=

180m 180 m 1 min = ×

1min 1min 60s

= 3 m s-1

Average velocity =

Displacement

Totaltimetaken

=

0m

60 s

= 0 m s–1

The average speed of Usha is 3 m s–1

and her average velocity is 0 m s–1.

8.3 Rate of Change of Velocity

During uniform motion of an object along a straight line, the velocity remains constant with time. In this case, the change in velocity of the object for any time interval is zero. However, in non-uniform motion, velocity varies with time. It has different values at different instants and at different points of the path. Thus, the change in velocity of the object during any time interval is not zero. Can we now express the change in velocity of an object?

Activity ______________ 8.7

• At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning.

• Can you answer why this happens?

• Measure this time interval using a digital wrist watch or a stop watch.

• Calculate the distance of the nearest point of lightning. (Speed of sound in air = 346 m s-1.) uestions

1. Distinguish between speed and velocity. 2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? 3. What does the odometer of an automobile measure? 4. What does the path of an object look like when it is in uniform motion? 5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108

m s–1.

Example 8.2 The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 h, calculate the average speed of the car in km h–1 and m s–1.

Solution:

Distance covered by the car, s = 2400 km – 2000 km = 400 km Time elapsed, t = 8 h Average speed of the car is,

vav = 400 km

8 h = s

t

= 50 km h–1

Q

2020-21

Activity 8.7 1000m At a time when it is cloudv. there may sound of lhLi11uer takes Lil to reach you after see the rtnlng. Cali VUu answer uljs hauucnS? Measure thls time interval ueing , wrist watch or a stoo waLCiL. te the distance of the nearest pomt 01 [Speed of sounu in uestions 50 h The average speed of the car is 0.0 Usma SWLLUS In a Long puul. Slie covers 180 m in one minuu-e by swimming from one end to the other aptl ba,0K altn the same stra•i Eiht path. Find the average speed ana average velocity of Usna. Tutal distance covered by Ush' 1 min Displacement of Usha in 1 •ain = O m l. 8. mogntüde an. eqvrÄ. does $it a signal the " minutes. d tance of the 180 m min I min 60 s min 3m s fro t the end o: 2400 kl tnp toc the ground traæls at is, 3 x tril . ff rne Average speed ot the car is, 50 km h- era, -e velocity = Total time taken 60 8.3 R Ite of Chai e of Velocity During uniform motion of an obieet along a nr Ime, the velocirv remams consrant nrne. In this case, the charwe in velocity of the object for anv tinme fiucrval is zero. However, in non-unhorm motion, velocity varies time. It has dj crent values at di it-rent inatants and at dill.erent points of tne path. Thus, the cnanee In velocity ot tne oblect dunng any tlme Inrervai is not zero. Can we now express me change in velocity of ScrmvCE uestions l. 8. olution: and baek Che same natl Find the average speed and averag distance covered by Ush' 1 mi Displacement of Usha in 1 •ain = O n mogntüde an. equ-A. it a signal the " minutes. d tance of the min 3m 1 ruin 60s fro at the star the ground traæls at is, trip and Average speed of the car is, era, -e velocity Lirne La«en 60 s —0m s-l The average speed of Usha is 3 m 83 F tte of of V During uniform motion of an object alon: str."Ä-11t line, the velocity remains const.% With time, this case, the cnenge in veloc or the omect tor any time inTerva1 is ze However, in non-unitorm mntjnn, vanes tirne. It has di werent values duacrent instants and at diner-ent poitll.s ##### Scene 6 (17m 55s) [Audio] MOTION 103 To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is, acceleration = change in velocity time taken If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, v – u a = t ( 8.3) This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2 . If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration. Activity ______________ 8.8 • In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. • Can you identify one example each for the above type of motion? Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s– 1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. Solution: In the first case: initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; time, t = 30 s . From Eq. (8.3), we have v – u a = t Substituting the given values of u, v and t in the above equation, we get ( ) –1 –1 6m s – 0m s = 30 s a = 0.2 m s–2 In the second case: initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s. Then, ( ) –1 –1 4m s – 6m s = 5 s a = –0.4 m s– 2 . The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is – 0.4 m s–2. uestions 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration? 2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus. 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration. Q 2020- 21 MOTION 103 To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is, acceleration = change in velocity time taken If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, v – u a = t (8.3) This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2 . If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration. Activity ______________ 8.8 • In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. • Can you identify one example each for the above type of motion? Example 8.4 Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. Solution: In the first case: initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; time, t = 30 s . From Eq. (8.3), we have v – u a = t Substituting the given values of u,v and t in the above equation, we get ( ) –1 –1 6m s – 0m s = 30 s a = 0.2 m s–2 In the second case: initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s. Then, ( ) –1 –1 4m s – 6m s = 5 s a = –0.4 m s–2 . The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is –0.4 m s–2. uestions 1. When will you say a body is in (i) uniform acceleration? (ii) non- uniform acceleration? 2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus. 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration. Q 2020-21 To answer such a question, we have to introduce anotnpr physical quantity called arceieration, wnich is measure of the chanue in the velocity of an object per unit change in velocity acceleration = time taken if the velocity of an obiect changes from an initial value u to [he finai value v time t, the acceleration a is, This kjnd of motion is known as accelerated motion. The, acceleration is taken to be posdive ii iL iS in the direction oiveiocny and negative when it is ooposite to the direction of velocity. The SI unit of It an orneer travels in a straiøht line and its velocity Increases or decreases by eaual amounus in equal intervals of Lillie. then i .e acceleration of the object is *Gid to unitorm. "Ihe motion ot a freely tällir g body is an example ot unitarmly acne-I •rate I motion. (m lhe other hand, an 0b can travel non-uniibrm acceie, if its velocitv chanees at a non-uniform :ate. For increases its speed lal amr •ts in equal Intervals 01 Time. tne car i. sav; to be moving wun non-unixorm 'lei Activity 8.8 (a) acceleratk • . is In the directlon of motion, (b) accele •atl is against the die -l motion, (c) ac. elen is unitorrn. (dl acce. :raLion is nun-uuiforrn. Can you Idenl"fy one exanple each for the above type Ot motion? puz.uon, Rahul padåles his bicycle to Monon attain a velocity of 6 m s-l in 30 s. Then he brakes such that the veloetty OJ me Dicycle comes (Inwn to 4 m s-l in the. next s. Cmcu•ate me acceleration of the bicycle in both the cases. initial ve'or•ity, u = O . From Eq. (8.3), we have S"bqfituting the given valu o: u,vand tin cne above equaLior, iritial •loc ty, u = 6m s 'Ihe accelerai.ion of the bicvcle in the first case is 0.2 m s-9 and in the second uestions vu:rs- fa. bes €4 km. trL h-: in S staring speed. the acceleration a is, This kind of motion is known as accelerated motion. The accelerption is t*ikpn to De postrtve in me an-ecnon ot veinetty and neeacive when it is opposite to the directiun of velocity. The SI unit of accelerai.ion is m s-2 an oblect travels in a straitfht line and its velocity increases or decrea«es by equal amounts in equal Intervnls ot nrne, then v .e accelera Lion Of the oblect is said to uniiorm. The motion of a freely fallin body is an example of uniformly acoel •rate I motion. On the other hand, an 0b, can travel wjrh non-unnorm accele, if Its veiociTV chanpes at a nnn-ururnrm -'ate. For exarno/e, if a car Craven' II", a stra_iv.lt roau Hlereases its • in equal of dime, Gw-n Lhe car i. to be moving with non-uniiorm 'lei t.lon Activity a range of munons in . Ch 8.8 (a) acre]c-ratie. is In the directlon of (b) accele •atl is against . motion, (c) ac lion is unilorrn. (d) acce. -'ration is non-uniinrm. Can you Identify one example each From Eq. (8.3), we have SubstitutinQ the Qiven valu 's o: u, c tm tne above equauor, •loc cy, u = 6 m s-l final •looty, u = 4 m s-l 5s arreleraTion of the hlovele in first case is 0.2 m and in lhe uestions fa. bes €4 km. trL h-l in. staring ##### Scene 7 (21m 49s) [Audio] SCIENCE 104 8.4 Graphical Representation of Motion Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables. To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time. 8. 4.1 DISTANCE–TIME GRAPHS The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x– axis and distance is taken along the y-axis. Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest etc. Fig. 8.3: Distance-time graph of an object moving with uniform speed We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in Fig. 8.3. The portion OB of the graph shows that the distance is increasing at a uniform rate. Note that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis. We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Fig 8.3. Draw a line parallel to the x-axis from point A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval (t2 – t1) while BC corresponds to the distance (s2 – s1). We can see from the graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t2 – t1). The speed, v of the object, therefore can be represented as v = 2 1 2 1 – – s s t t ( 8.4) We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the distance travelled by a car in a time interval of two seconds. Table 8.2: Distance travelled by a car at regular time intervals Time in seconds Distance in metres 0 0 2 1 4 4 6 9 8 16 10 25 12 36 2020- 21 SCIENCE 104 8.4 Graphical Representation of Motion Graphs provide a convenient method to present basic information about a variety of events. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over. As you have studied in mathematics, a straight line graph helps in solving a linear equation having two variables. To describe the motion of an object, we can use line graphs. In this case, line graphs show dependence of one physical quantity, such as distance or velocity, on another quantity, such as time. 8.4.1 DISTANCE–TIME GRAPHS The change in the position of an object with time can be represented on the distance-time graph adopting a convenient scale of choice. In this graph, time is taken along the x–axis and distance is taken along the y-axis. Distance-time graphs can be employed under various conditions where objects move with uniform speed, non-uniform speed, remain at rest etc. Fig. 8.3: Distance-time graph of an object moving with uniform speed We know that when an object travels equal distances in equal intervals of time, it moves with uniform speed. This shows that the distance travelled by the object is directly proportional to time taken. Thus, for uniform speed, a graph of distance travelled against time is a straight line, as shown in Fig. 8.3. The portion OB of the graph shows that the distance is increasing at a uniform rate. Note that, you can also use the term uniform velocity in place of uniform speed if you take the magnitude of displacement equal to the distance travelled by the object along the y-axis. We can use the distance-time graph to determine the speed of an object. To do so, consider a small part AB of the distance-time graph shown in Fig 8.3. Draw a line parallel to the x-axis from point A and another line parallel to the y-axis from point B. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval (t2 – t1) while BC corresponds to the distance (s2 – s1). We can see from the graph that as the object moves from the point A to B, it covers a distance (s2 – s1) in time (t2 – t1). The speed, v of the object, therefore can be represented as v = 2 1 2 1 s s t t (8.4) We can also plot the distance-time graph for accelerated motion. Table 8.2 shows the distance travelled by a car in a time interval of two seconds. Table 8.2: Distance travelled by a car at regular time intervals Time in seconds Distance in metres 0 0 2 1 4 4 6 9 8 16 10 25 12 36 2020-21 8.4 1 :cal of Graohs provide a convenient method to prcsent b.] 'öic inform-anon about a vaciety of events. For example, in the telecast ot a one-day cricket maT0h, vertical bar graphs show the run rate of a team in each over, As you have studied in '_:lucn.a€ies. a strai[iht h. in sol',iug a linear equacion having two variames. To describe the motion of an ohiect, we can use line urauhs. In this case. line uraons show of one physical quami€,i.v, such as distance or velocity, on another quantity, such as time. 8.4.1 DISTANCE-TIME GRAPHS me charge in the position of an ohiect with time can be represented on the distance-time a convenient scale chntre. In VI-don, Orne is t_dKeE1 alone the Y—axis and distance is taken a10i1Q the y-axis various I-ions where ohiects move ununrrn speed, non-unnonn speed, .-en- 10 o Fig. 8.3: Di*$ance-tt. •e groph of an object moving wetn unyorm speed We know that when an ohipc•t travels eanal distances in equal illtervais 01 time, it moves urgilorrn speed. This shows that the die,tance travQ]ed by the ohiect is directly prnnoruonal co rune Taken. for sueed, a eraDh of distance travelled ayeu11St The portion ot the graph shows that the distance is increasjng at a 1 mnuorrn rate. Note that, you can atso use the term uniiorm veiocuv in Oiaee of soeed if vou take the of equal to the dlQtance travelled by tne Object along tne we can use the distance-time granh to determine the speed of an obiect. To do so, consiEIer a small part. AB of the distence-Cime gr:ann snoum mg 8.3. Draw a line to rne x-axis from point A and ann her line parallel to [he y-axis irom point B. The -e two the rune interval [t — t,) w, le corresnonds to tb' distance {s? — S I, •an see from [he graoh LAa€as the ves frum the point c-,-n also plot the diqtance-time graph '01 elera_tpql motion. ;le. 8.2 shows the travel lea DY a car in a time interval of two seconds. e 8.2: D stance trave e y car at regular time intervals Time in seconds Distance in metres 10 12 16 25 36 ScrmvCE

vine two variables. "l"o describe the motion of an obtect, we use Grnnns. depenuence Of one pnvslcal quanu1V, ch as disuance or velocity, on anotuer antity, such as time. Ll DISTANCE-TIME GRAPHS e change. in the position of an ohiect with can be reoreserued on the Giseanee-time a convenient scale of choice. ume is taken alone Ehe y-axis d dlsranc:e iS taxen along rne y-ar•s stanee-U_Lne vraons can be ernoloved riuus witere obiects move speed, non-unitonn speed, .-en- 8.3: Distance-h. •e aranh of an object moving We can use the distance-time graph determine the, speed ot an oblect. To do cumsider a small oart AB of the distance-til g, shown in Fik 8.3. Draw a lir•e para. to the x-axis trom point A and ann her Ij parallel to rhp y-ans rrom point B. _e t lines meet each 0Lner at Poir L to ABC. NOW, on thc denu the Unje interval (t w, le L Q correspon to th' distance (52 — s i', We •an see trom grant. rhe ves Irom Vhe po tt B. iL covers db. •ance (sm— S,) in v the object, theref( W also plot the distance-time gra C *stance trav tllcd by a car in a time inter ot two seconds. car at re ular time intervals Time in seconds Distance in metre

##### Scene 8 (25m 14s)

[Audio] MOTION 105 The distance-time graph for the motion of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier distance-time graph ( Fig. 8.3) for uniform motion. The nature of this graph shows nonlinear variation of the distance travelled by the car with time. Thus, the graph shown in Fig 8.4 represents motion with non-uniform speed. 8. 4. 2 VELOCITY-TIME GRAPHS The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis Fig. 8.4: Distance-time graph for a car moving with non-uniform speed Fig. 8.5: Velocity-time graph for uniform motion of a car and the velocity is represented along the y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time (Fig. 8.5). It will be a straight line parallel to the x-axis. Fig. 8.5 shows the velocity-time graph for a car moving with uniform velocity of 40 km h–1. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement. To know the distance moved by the car between time t1 and t2 using Fig. 8.5, draw perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of 40 km h–1 is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB. So, the distance s moved by the car in time (t2 – t1) can be expressed as s = AC × CD = [( 40 km h–1) × (t2 – t1) h] = 40 (t2– t1) km = area of the rectangle ABDC (shaded in Fig. 8.5). We can also study about uniformly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car, in km h– 1 as well as in m s–1, at different instants of time is shown in table 8.3. Table 8.3: Velocity of a car at regular instants of time Time Velocity of the car (s) (m s–1) (km h –1) 0 0 0 5 2.5 9 10 5.0 18 15 7.5 27 20 10.0 36 25 12.5 45 30 15.0 54 2020- 21

MOTION 105

The distance-time graph for the motion of the car is shown in Fig. 8.4. Note that the shape of this graph is different from the earlier distance-time graph (Fig. 8.3) for uniform motion. The nature of this graph shows non- linear variation of the distance travelled by the car with time. Thus, the graph shown in Fig 8.4 represents motion with non-uniform speed.

8.4.2 VELOCITY-TIME GRAPHS

The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the x-axis

Fig. 8.4: Distance-time graph for a car moving with non-uniform speed

Fig. 8.5: Velocity-time graph for uniform motion of a car

and the velocity is represented along the y-axis. If the object moves at uniform velocity, the height of its velocity-time graph will not change with time (Fig. 8.5). It will be a straight line parallel to the x-axis. Fig. 8.5 shows the velocity-time graph for a car moving with uniform velocity of 40 km h–1. We know that the product of velocity and time give displacement of an object moving with uniform velocity. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement. To know the distance moved by the car between time t1 and t2 using Fig. 8.5, draw perpendiculars from the points corresponding to the time t1 and t2 on the graph. The velocity of 40 km h–1 is represented by the height AC or BD and the time (t2 – t1) is represented by the length AB. So, the distance s moved by the car in time (t2 – t1) can be expressed as s = AC × CD = [(40 km h–1) × (t2 – t1) h] = 40 (t2– t1) km = area of the rectangle ABDC (shaded in Fig. 8.5). We can also study about uniformly accelerated motion by plotting its velocity– time graph. Consider a car being driven along a straight road for testing its engine. Suppose a person sitting next to the driver records its velocity after every 5 seconds by noting the reading of the speedometer of the car. The velocity of the car, in km h–1 as well as in m s–1, at different instants of time is shown in table 8.3.

Table 8.3: Velocity of a car at regular instants of time

Time Velocity of the car (s) (m s–1) (km h –1)

0 0 0 5 2.5 9 10 5.0 18 15 7.5 27 20 10.0 36 25 12.5 45 30 15.0 54

2020-21

35 Fig. 8.4: Distance-time aranhfor a car moving with The distance-time graph for the motion of the car is shown in 2, 24.4. Nnt_e Shane: rnis JS €11 tne disLance-tuue graph (k 8.3) for u Luorn_ The neoure of this shm s nor- lilltear varia at' the disiuuee tra. ell. oy tne ear time. Thus, tne in Fip 8.4 represents monon witn n0i. -unnorm The variation in velocity with G. T (e •r an ohlect moving in a stratdtlt line "-an be renresenred bv a veiooitv-urv In this graph, tune is represen'0J3 tile x-axis Time (h) • . : Velæity-ärne graphfor uniform rnotion q/ mg 8.5 Monon and the velocity is represented along the rhe onlec•t moves ururnrm the he121it oi iLS velocnv-L'i11me not time Q. 8.5). It a stra. , : velocity-tjme graph tör a car moving witn urn101-m Velocir.v Of 40 km h-l We know that the muduct of velocity and tiple of an obic.ct moviu.g tu„lttorrn veqmtty, area eneJ0€æd by veinetry-ume grann and tne: ome axts Wili ne equal to the rnaelütuue 01 tile disoLaeemer1t. To know the disiance moved bv the car bet*.repn time t, and t, Fig. 8.5, dr•s.w the crxrregponrtlng to r ne Time and on The pranh "he veloeuy of 40 km h-L is revresentcd DV he AC or BD and the thne (G— 4) •s cpacsen€ccl by So, the distance mm •d by the car in tir .e (ta — tl) can be r xpre. -sed as arec of he rectangle ABDC (shaded also studv ahout uniformly acc (ert '-ed motion by .g its ve-100ity- ann. Consu-lera car nerno driven along - St' dll_• FLL road lor cesüné its en21_ne. SuoDose a pet-sun next to die driver recu,d3 its ot tne sneedometer ot tne car. 'lhe ve•oc1W of the, car, in km h-l as well as in m s-l, at rerent insi_alits of time is shown in table 8.3. re ular instants of time Time (s) 10 15 20 25 30 Velocity of the car 2.5 5.0 7.5 10.0 12.5 15.0 (km KI) 18 27 36 45 54

10 o Fig. 8.4: Distance-firrte a car moving with n01t-Lutyorrrt swed The distance-time m-anh for the motion of the car is shown in Q. 8.4. Note tl (listanee-ttrne graph (Wig. 8.3) tor v Illorn_ rnnrnnn. The nature Of oranh shm nor hncar valiuuun of the tra. oy che car time. 'Illus. the FÄg 8.4 represents motion wiLh noi. -unilorm 8.4.2 VELOCITY-TIME The variation in velocity with t. •r an oblect movinz in a strai2i1t line "-an be represented a velocity-tim s, •aph, In this equal to the m:" 'it"de of the To know the disrance movea ny rl« between time t, aud t, usitlé Fi±. 8.3. perpendir-ulars from the poini.s correspu to the time and t2 on the s'raph 'he ve ot 1 km h-l is representea Dy he or BD and the time (ta— 4) 's the So, the dista_nce s mm •d by the ( • (t40 kn. x (t, —t) arec of he rectangle ABDC (sl a.so study about unifc acc lert '-ed moLion by its vel• - st- ail Flt road tor testing its ev ftine. Suj a person next (inver reenr veLucWv aiccr cverv 5 secunds bv notin reaG, of the speedome€er of the cal veloc:ty or the ear, in kill as well m s-l, at cil"erent instants of time IS s in table 8.3. e oc yo a cara gular instants of time Velocity of the car Time (s) 10 2.5 5.0 (km h - 18

##### Scene 9 (28m 41s)

[Audio] SCIENCE 106 In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line. Fig. 8.6: Velocity-time graph for a car moving with uniform accelerations. You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance ( magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph (Fig. 8.6). Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph (Fig. 8.6). That is, s = area ABCDE = area of the rectangle ABCD + area of the triangle ADE = AB × BC + 1 2 (AD × DE) In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape. Fig. 8.7: Velocity-time graphs of an object in nonuniformly accelerated motion. Fig. 8.7(a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Fig. 8.7 (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Try to interpret these graphs. Activity ______________ 8.9 • The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table 8.4. Table 8.4: Distances of stations B and C from A and times of arrival and departure of the train Station Distance Time of Time of from A arrival departure (km) (hours) (hours) A 0 08: 00 08: 15 B 120 11:15 11: 30 C 180 13:00 13:15 • Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform. Velocity (km h –1) 2020- 21

SCIENCE 106

In this case, the velocity-time graph for the motion of the car is shown in Fig. 8.6. The nature of the graph shows that velocity changes by equal amounts in equal intervals of time. Thus, for all uniformly accelerated motion, the velocity-time graph is a straight line.

Fig. 8.6: Velocity-time graph for a car moving with uniform accelerations.

You can also determine the distance moved by the car from its velocity-time graph. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph (Fig. 8.6). Since the magnitude of the velocity of the car is changing due to acceleration, the distance s travelled by the car will be given by the area ABCDE under the velocity-time graph (Fig. 8.6). That is, s = area ABCDE = area of the rectangle ABCD + area of the triangle ADE

= AB × BC +

1 2 (AD × DE)

In the case of non-uniformly accelerated motion, velocity-time graphs can have any shape.

Fig. 8.7: Velocity-time graphs of an object in non- uniformly accelerated motion.

Fig. 8.7(a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time while Fig. 8.7 (b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time. Try to interpret these graphs.

Activity ______________ 8.9

• The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table 8.4.

Table 8.4: Distances of stations B and C from A and times of arrival and departure of the train

Station Distance Time of Time of from A arrival departure (km) (hours) (hours)

A 0 08:00 08:15 B 120 11:15 11:30 C 180 13:00 13:15

• Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.

Velocity (km h –1)

2020-21

In this case, the velry•jty-time graph for Cbe monon ot rne car is snnwn 111 too. 8.6. Toe g 2r nature of the graoh shows tuat velocity clu bv equal uluounts in equal intervals ot Thus, tor all accelerated motton, tne velocity-time graph is a straignt line t} 10 Fig. 8.6: Velrrit!'-time grrwh for a car moving with You can also determine the diqt* nce the ear from its velocity-time The area under the velocity-time the di•.tance of di:4] •ment) If tne car would have bee ' m with mmilorm velocrtv. the distaxre vruvehed b would be represented by the area maonitude or tne vmocjty of the ear is charip]np due. to acceleration. t' -stance s bv the car wdl be 'en the area ATÆCLFE under the velocit) -t graph area ABCDI area of i -ecaangle ABCD + area of 2 In the case of non-uniformly accelerated motion, velocity-time graphs can have any 10 sc ec 3 4 Time (s) 5 o Fig. 8 Velryy:ty-ttme granhs of an object tn non- unyonruy a •cet •t motton. *ig. 713) sb. •vs .rplncitv-time tx leureserts n. •tiun 01 ail ut.)leei, wnuse city is ore with ti±ne waile reoresenti-u T ne. non-unllorm Variation of velucih ot object with Lime. Try Activity 8.9 e me. times of arrival and denariure of a train at three. A, B and C and the cilsrance_ Qi sranong B C from s ances o s a Ions and C from A and times of arriv d departure Station Distance from A 120 180 th trai Time of Time of arrival departure (hours) 08:00 11:15 13:00 (hours) 08:15 1 1:30 13:15 F\at and interpret t.ne dietance-tjme for the train assu111J112 that its motmn between any two stations is uniaorm. ScrmvCE

1. 8.6: Velocity-time graph for a car moving with uregörrrt accelerations. You can also determine the disü nce ved bv the car trom its velority-time e area unaer the ve1001tv-nrnp Q, •w. distance (inauL1iLude Of •ment) wed bv Lhe car in a kiv-en of chne. the ear would have bee 11 m w:lth veloejty, the 'Ijstali?e VT D' uder the graph (F W. 8.6). Sin,e t he IQi1itude of the velocity of the ear is due to acceleration, t' Le 'stance s a oy the car oe tne area under the velocnyt:me graph Q. 8.6). area or r te -ee.nngle ABCD + area of 8 Fig. 8 7; Velocttu-ttme aranhs of an object tn a •cet _ motion. reltreserrs n. OT an ontect prelu Cu y is a, ore Ising wit h time wh F 2. 8.7 (b) s'•ows the veloeitv-tiille era f he non-unitorm varial ion or . ae oojectwlth ume. Try to Interp Activity 8.9 1 me times Of arrival and departure of a train at l_nree sEanons A, b and C and i and C from A and times of arriv' and de arture of the train Station Distance from A (km) 120 180 Time of Time o arrival departu (hours) 08:00 11:15 13:00 (hours) 08:15 1 1:30 13:15

##### Scene 10 (31m 33s)

[Audio] MOTION 107 8.5 Equations of Motion by Graphical Method When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below: v = u + at (8.5) s = ut + ½ at2 ( 8.6) 2 a s = v2 – u2 (8.7) where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation. Eq. ( 8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method. 8. 5.1 EQUATION FOR VELOCITY-TIME RELATION Consider the velocity-time graph of an object that moves under uniform acceleration as Activity _____________ 8.10 • Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 8.5 shows the distance travelled by them in different times Table 8.5: Distance covered by Feroz and Sania at different times on their bicycles Time Distance Distance travelled travelled by Feroz by Sania (km) (km) 8: 00 am 0 0 8: 05 am 1.0 0.8 8: 10 am 1.9 1.6 8: 15 am 2.8 2.3 8: 20 am 3.6 3.0 8: 25 am – 3.6 Q • Plot the distance-time graph for their motions on the same scale and interpret. uestions 1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? 3. What can you say about the motion of an object if its speedtime graph is a straight line parallel to the time axis? 4. What is the quantity which is measured by the area occupied below the velocity-time graph? Fig. 8.8: Velocity-time graph to obtain the equations of motion 2020- 21

MOTION 107

8.5 Equations of Motion by Graphical Method

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations are given below: v = u + at (8.5) s = ut + ½ at2 (8.6) 2 a s = v2 – u2 (8.7) where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (8.5) describes the velocity-time relation and Eq. (8.6) represents the position-time relation. Eq. (8.7), which represents the relation between the position and the velocity, can be obtained from Eqs. (8.5) and (8.6) by eliminating t. These three equations can be derived by graphical method.

8.5.1 EQUATION FOR VELOCITY-TIME

RELATION

Consider the velocity-time graph of an object that moves under uniform acceleration as

Activity _____________8.10

• Feroz and his sister Sania go to school on their bicycles. Both of them start at the same time from their home but take different times to reach the school although they follow the same route. Table 8.5 shows the distance travelled by them in different times

Table 8.5: Distance covered by Feroz and Sania at different times on their bicycles

Time Distance Distance travelled travelled by Feroz by Sania (km) (km)

8:00 am 0 0

8:05 am 1.0 0.8

8:10 am 1.9 1.6

8:15 am 2.8 2.3

8:20 am 3.6 3.0

8:25 am – 3.6

Q

• Plot the distance-time graph for their motions on the same scale and interpret.

uestions

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? 2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis? 3. What can you say about the motion of an object if its speed- time graph is a straight line parallel to the time axis? 4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Fig. 8.8: Velocity-time graph to obtain the equations of motion

2020-21

Activity 8.10 Feroz and his sister Snni.o go to sphool on rnelr meveles, tsnrh 01 rnem srarr. at the same tune from thuir holile but take nrerent. runes to reach r.ne sennol H.5 rue 'iterance Eravened able 8.5: Distance covered by Fero and Sania at different times o their bicycles 0.8 1.6 2.3 3.0 3.6 Time 8:00 am 8:05 am 8:10 am 8:15 am 8:20 am 8:25 am Distance travelled by Feroz (km) 1.0 1.9 2.8 3.6 Distance travelled by Sania (km) Plot the distance-time graph for tneir motions on the same scale and uestions l. 4. of the graphsfor unconn and non-un.Vorm rnotion of What can you say of an object dismrzce-tfrne une parallel What can goo, nuåon Qf true occtu.péc•d, veioci*th'-f\$ne 8.5 Equations of hCotion by When an obiect moves along a straight line relate its veinejty, acceleration dn_ring motion anci Lhe ciisv_ance covered bv it in a ceruain tirne interval by a set of ations known as the era 'Atlons of motion. For conventpnce, a set oi unree such equauons are given below: where u is the initial velocitv of the obiect w] lich moves with utij tnrm acceleration ator distance travelled by the 2t i- tin.e t Eq. (8.5) describes the velo -tune rela tion ann EA. t8.5J represenr= e •,osjrjon-tlme retia Lion. Eq. (8.7), v lich. reoresenLS tile rel the ,.JOSL veloci irnir..ktinp t. 'aes t.nree. equauons can be 8.5.1 EQUA'i10N FOR VELOCITY-TIME the velocity-time Qraoh of an obiect moves unüer uniiui'ill accelerauon as Time t,) Fig. 8.8: graph to obtain the equations Monon

Time 8:00 am 8:05 am 8:10 am 8:15 am 8:20 am 8:25 am ana sama at their bicycles 0.8 1.6 2.3 3.0 3.6 amerent times o Distance Distance travelled travelled by Feroz by Sania (km) (km) 1.0 1.9 2.8 3.6 where u is the initial velncitv of the tittie t, V is the fif].11 s travelled by t,ne i. Eq. lb.5) deserwes the velo -tune re: and Eq. (8.6) represent= e -.,osN.ion re' anon t)eiureen the •on Ineve] can he obtainea fro. = B, (S, 5) and uenvea memod. 8.5.1 ION FOR VELOCITY- the velocity-time graph of an Inuves unaer unnorlll acceleraUl Hot tne distance-time graph "or tneir motions on the saule scale arid uestions l. of the graphsfor unconn and non-un.Vorm rnotion of What can you say of an object dismrzce-tfrne une parallel What can goo, nution Qf

##### Scene 11 (34m 36s)

[Audio] SCIENCE 108 = OA × OC + 1 2 (AD × BD) ( 8.10) Substituting OA = u, OC = AD = t and BD = at, we get s = u × t + 1 ( ) 2 t ×at or s = u t + 1 2 a t 2 8.5. 3 EQUATION FOR POSITION– VELOCITY RELATION From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is, s = area of the trapezium OABC = ( ) OA + BC ×OC 2 Substituting OA = u, BC = v and OC = t, we get ( ) 2 = u + v t s ( 8.11) From the velocity-time relation ( Eq. 8.6), we get ( ) v – u t = a (8.12) Using Eqs. (8.11) and ( 8.12) we have ( ) ( ) × v + u v - u s= 2a or 2 a s = v2 – u2 Example 8.5 A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity. shown in Fig. 8.8 (similar to Fig. 8.6, but now with u ≠ 0). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t. Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA Substituting BC = v and OA = u, we get v = BD + u or BD = v – u (8.8) From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by a = Change in velocity time taken = BD BD = AD OC Substituting OC = t, we get a = BD t or BD = at (8.9) Using Eqs. ( 8.8) and ( 8.9) we get v = u + at 8. 5.2 EQUATION FOR POSITION-TIME RELATION Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distance s travelled by the object is given by s = area OABC (which is a trapezium) = area of the rectangle OADC + area of the triangle ABD 2020- 21

SCIENCE 108

= OA × OC +

1 2 (AD × BD) (8.10)

Substituting OA = u, OC = AD = t and BD = at, we get

s = u × t +

1 ( )

2

t ×at

or s = u t + 1

2

a t 2

8.5.3 EQUATION FOR POSITION–VELOCITY

RELATION

From the velocity-time graph shown in Fig. 8.8, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is, s = area of the trapezium OABC

= ( ) OA + BC ×OC

2

Substituting OA = u, BC = v and OC = t, we get

( )

2

=

u + v t

s (8.11)

From the velocity-time relation (Eq. 8.6), we get

( ) v – u t = a (8.12)

Using Eqs. (8.11) and (8.12) we have

( ) ( ) × v + u v - u s= 2a

or 2 a s = v2 – u2

Example 8.5 A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity.

shown in Fig. 8.8 (similar to Fig. 8.6, but now with u ≠ 0). From this graph, you can see that initial velocity of the object is u (at point A) and then it increases to v (at point B) in time t. The velocity changes at a uniform rate a. In Fig. 8.8, the perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t. Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA Substituting BC = v and OA = u, we get v = BD + u or BD = v – u (8.8) From the velocity-time graph (Fig. 8.8), the acceleration of the object is given by

a =

Change in velocity

time taken

=

BD BD =

Substituting OC = t, we get

a =

BD

t

or BD = at (8.9)

Using Eqs. (8.8) and (8.9) we get v = u + at

8.5.2 EQUATION FOR POSITION-TIME RELATION

Let us consider that the object has travelled a distance s in time t under uniform acceleration a. In Fig. 8.8, the distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB.

Thus, the distance s travelled by the object is given by

s = area OABC (which is a trapezium) = area of the rectangle OADC + area of the triangle ABD

2020-21

shown in Fig, 8.8 (eirnilar to Fio. 8.6, but now From ernnh, you can See 'hat iruual velocitv of the oi_neet is u (at bornt A) then it increases to v in tilne t- The velocity at a rate a. Ftff. 8.8, the pernentljcular lines BC and BE, are drawn from pomt B on tne time and the velocity axes resvectivelv, so that the initial velueity is represc.nted by ON the filial vevv•itv iS by BC and tne ne inrerval t is renre.senred by OC. HD = BC • CD. represents the change in velocity in time Let us draw AD parallel to OC. From the Subsl:lkutiug BC = v and OA : (8.8) or From the velociLv-tixne graph (Fig. 8.8). the acceleration of the oleieet is given by AD OC Substituting OC = t, we get = OAX OC + (AD x BD) Substituting OA = u, OC = AD = t and BD = at 2 2 8.5.3 EQUATION FOR POSITION-VELOCITY From the velocitv-time graph shown in FiQ. 8.8, the distance s travelk.d b: obiect in time t, moving under unitorrz „-ration a is given by the area enc10k v itmn tne trapezium OAHU unner e •apn. 'Inat is, 2 Substituting •rcnaa 2 the velocity-time relation (Eq. 8.6), or BD = at Using Eqs. (8.8) and (8,9) we 8.5.2 EQUATION FOR POSITION-UME Let us consider that the ohiec, has travelled distance in timr„ t under uniiorm acceleration a, In c.d, the distance enmosea •tvnnm : unaer tne velocity-time Thus, the distance s travelled by the object = area OARC (which is a trapezium) area of Lhe rectangle OÄDC + alea of the Using Eqs. (8.11) and (8.12) we have attains a velocity of 72 km h-l in 5 minutes. Assuming that the acceleration is unitnrm. find K) the acceleratinn and (li) the djstanee travelled by the train for attaining this veiociuy ScrmvCE

represents the change in velocity in time Let us draw AD Darallel to OC. From the SuosLILuting BC v and OA = or Fram the veloc1TV-time graph (F irk. 8.5). acceleration of the ooject is given by Chan r. Ln velocity AD OC bstituting OC = t, we get BD = at ing Eqs. (8.8) and (8,9) we 5.2 EQUATION FOR POSITION-UME t us consider that the oblec, has travelled under unitorm (l*stance s in timr„ oeierarton a, In c.d, the distance bv the oi is 'bl.ained bv che arca closed wooin •AL n. I-unlel tile velocity-tnne Thus, the distance s travelled by the object 8.5.3 EQUATION FOR POSITION-VELOCI From the velocity-time grnnh shown F 12.8.8, the disvance s Lraveueu in Lime t, movinu under unhurr- --raLi a is eiven bv the area enclu Eii.liin trapezaum OABC under 'V e k •avn. Tl'tat i: (8. 2 the velocity-time relation (Eq. 8. (8. Using Eqs. (8.11) and (8.12) we have s=— or Lxalllnne 0.0 star L Lile mroru re: atLåins velocity of 72 km h-l 5 minutes. Assuming that th

##### Scene 12 (38m 20s)

[Audio] MOTION 109 Solution: We have been given u = 0 ; v = 72 km h–1 = 20 m s-1 and t = 5 minutes = 300 s. (i) From Eq. (8.5) we know that ( ) v – u a = t –1 –1 –2 20 m s – 0 m s = 300s 1 = m s 15 (ii) From Eq. ( 8.7) we have 2 a s = v2 – u2 = v2 – 0 Thus, 2 –1 2 –2 = 2 (20 m s ) = 2×( 1/15) m s v s a = 3000 m = 3 km The acceleration of the train is 1 15 m s– 2 and the distance travelled is 3 km. Example 8.6 A car accelerates uniformly from 18 km h –1 to 36 km h –1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time. Solution: We are given that u = 18 km h–1 = 5 m s–1 v = 36 km h–1 = 10 m s–1 and t = 5 s . (i) From Eq. (8.5) we have v – u a = t = -1 -1 10 m s – 5 m s 5s = 1 m s–2 (ii) From Eq. ( 8.6) we have s = u t + 1 2 a t 2 = 5 m s–1 × 5 s + 1 2 × 1 m s–2 × (5 s)2 = 25 m + 12.5 m = 37.5 m The acceleration of the car is 1 m s –2 and the distance covered is 37.5 m. Example 8.7 The brakes applied to a car produce an acceleration of 6 m s- 2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. Solution: We have been given a = – 6 m s–2 ; t = 2 s and v = 0 m s–1. From Eq. ( 8.5) we know that v = u + at 0 = u + (– 6 m s–2) × 2 s or u = 12 m s–1 . From Eq. (8.6) we get s = u t + 1 2 a t 2 = (12 m s–1 ) × (2 s) + 1 2 (– 6 m s–2 ) (2 s)2 = 24 m – 12 m = 12 m Thus, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? uestions 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. Q 2020- 21

MOTION 109 Solution: We have been given u = 0 ; v = 72 km h–1 = 20 m s-1 and t = 5 minutes = 300 s. (i) From Eq. (8.5) we know that ( ) v – u a = t –1 –1 –2 20 m s – 0 m s = 300s 1 = m s 15 (ii) From Eq. (8.7) we have 2 a s = v2 – u2 = v2 – 0 Thus, 2 –1 2 –2 = 2 (20 m s ) = 2×(1/15) m s v s a = 3000 m = 3 km The acceleration of the train is 1 15 m s– 2 and the distance travelled is 3 km. Example 8.6 A car accelerates uniformly from 18 km h –1 to 36 km h –1 in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time. Solution: We are given that u = 18 km h–1 = 5 m s–1 v = 36 km h–1 = 10 m s–1 and t = 5 s . (i) From Eq. (8.5) we have v – u a = t = -1 -1 10 m s – 5 m s 5s = 1 m s–2 (ii) From Eq. (8.6) we have s = u t + 1 2 a t 2 = 5 m s–1 × 5 s + 1 2 × 1 m s–2 × (5 s)2 = 25 m + 12.5 m = 37.5 m The acceleration of the car is 1 m s –2 and the distance covered is 37.5 m. Example 8.7 The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. Solution: We have been given a = – 6 m s–2 ; t = 2 s and v = 0 m s–1. From Eq. (8.5) we know that v = u + at 0 = u + (– 6 m s–2) × 2 s or u = 12 m s–1 . From Eq. (8.6) we get s = u t + 1 2 a t 2 = (12 m s–1 ) × (2 s) + 1 2 (–6 m s–2 ) (2 s)2 = 24 m – 12 m = 12 m Thus, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? uestions 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled. Q 2020-21

##### Scene 13 (41m 54s)

[Audio] SCIENCE 110 2. A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest. 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start? 4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start? 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s- 1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there? 8.6 Uniform Circular Motion When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion? Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD. Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners. How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times. Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b). In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)? It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely? If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point. If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion. We know that the circumference of a circle of radius r is given by π 2 r . If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by π 2 r v = t ( 8.13) When an object moves in a circular path with uniform speed, its motion is called uniform circular motion. (a) Rectangular track (b) Hexagonal track (d) A circular track (c) Octagonal shaped track Fig. 8.9: The motion of an athlete along closed tracks of different shapes. 2020- 21

SCIENCE 110

2. A train is travelling at a speed of 90 km h–1. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s-2. Find how far the train will go before it is brought to rest. 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s-2. What will be its velocity 3 s after the start? 4. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start? 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2

in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

8.6 Uniform Circular Motion

When the velocity of an object changes, we say that the object is accelerating. The change in the velocity could be due to change in its magnitude or the direction of the motion or both. Can you think of an example when an object does not change its magnitude of velocity but only its direction of motion?

Let us consider an example of the motion of a body along a closed path. Fig 8.9 (a) shows the path of an athlete along a rectangular track ABCD. Let us assume that the athlete runs at a uniform speed on the straight parts AB, BC, CD and DA of the track. In order to keep himself on track, he quickly changes his speed at the corners. How many times will the athlete have to change his direction of motion, while he completes one round? It is clear that to move in a rectangular track once, he has to change his direction of motion four times. Now, suppose instead of a rectangular track, the athlete is running along a hexagonal shaped path ABCDEF, as shown in Fig. 8.9(b). In this situation, the athlete will have to change his direction six times while he completes one round. What if the track was not a hexagon but a regular octagon, with eight equal sides as shown by ABCDEFGH in Fig. 8.9(c)? It is observed that as the number of sides of the track increases the athelete has to take turns more and more often. What would happen to the shape of the track as we go on increasing the number of sides indefinitely? If you do this you will notice that the shape of the track approaches the shape of a circle and the length of each of the sides will decrease to a point. If the athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. The motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion. We know that the circumference of a circle of radius r is given by π 2 r . If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by

π 2 r v =

t (8.13)

When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.

(a) Rectangular track (b) Hexagonal track

(d) A circular track (c) Octagonal shaped track

Fig. 8.9: The motion of an athlete along closed tracks of different shapes.

2020-21

3. 4. gféJfJ dcumwtx•d t,'uz When the velarltv of an ohiect chanoes, Sav [bat (_n_nect is aceelejaLi1iQ. Tue. eir _1iQe in the velocity coul'] be d' [e to cha its or the nn•ecnnn of the n or bolil. Can VOU think Of an e_Xd111Dle an olneet does not its de of velocity but ony its directiC,) .uouon? (a) Rectangular track (c) Octagonal shaped track (d] A circular track Fig. 8.9: mnHon Qfan athlete along closed tracks OJ dwuent shapes. Let us consider an example of the motion ofa body along a cloaed path. Fig 8.9 (a) snows the path ot an athlete a]nng a rectancfl'lar track Let us assnme tnat T.ne aTn•ece on the straiont parts AB. BC. Ci") and LIA or the trar•k. In orner to keep himself on track, he qurc:K1v chanpes his sneed at the corners. How mariv lirnes wili t-he atolete have to chanQe his direcuon of motion. while he conn.JieLes one round? It is clear that to move in a rectaneular track once, he has to change his direction of motion Now, suppose instead of a rectangular track, the athlete is running •long a hexagonal shaped path as hown in 8.9(b). In this situatic A, thc ath'ete will have to cha_nge his direc or six times wnlle he competes one noun. Wnat ir me traci was not _apm. but reQ"lar with eirlht 'ides as shown by jn R.S- It js observed that as "he. numbe - s Of the track increases thr• atheneLe : 'as ._aKe turns more and more Aten. •tld haouen to the shaDe of the tra . S we eo on uncreasinQ dle nulilber of in .1eiii1iLclv? If vou do diis vou will nc -j' .e tha the shave of the track av['roaches he of a circle and the lenr:2U1 of each of tk. sides will decrease to a point. If the ai hlete moves with a velocity ot constant 103 gnitude along the circular path, the only change in his velocity is due to the change in the direction ot motion. me morjon ot tne athlete movmg along a circular parh is, thererore, an examnl.e ot an accelerated motion. We know that tne ctrcumTerence of a circle of radius ris given by 21tr . It the athlete takes t seconds to 20 once around the circular path of radius r, the speed v is given by 21tr When an object moves in a circular path with nmrorm speed. ics motion is caned umiorrn circular motion. ScrmvCE

dcumwtx•d t,'uz 'len the veloe!ty of an chenges, the is Cn- Ille Velocii,v COULU be due to ii IKS or the direction of the n or tn. Can you 01 an an leet does not cnanoe. its mavr•irs de Of .ueity but only its dueetic,) . uution? (a) Rectangular track is tl'»at to rrl.nve in a once, he. has to change his direcuon of moti four times. Now, suppose instead of a rectangu track, the athlete is running • 1011* hexagonal shaped path ABCDER as ho in F 8.90)). In this situatic A, thL athl will have to change his direr; or six tin while he complelcs one nun. What if t tracl was not a he: -arm. but a reemx with eifht sides as shown . in Fig. R.SO? It is observed tt as •ne numm - or the track increas thr• atnelete : •as 'n turns more and ,gren. wnet uld nappen to the snane the tra . •s we go on Increasing rhe numi of sine. in If vou do cnis you nc -ire the shane. the, track ar»proael- 'he B: lane of a circle and lhe lern of eacr. tk. Slues will decrease LO a Donn. 11 Lhe athil moves with a veloeiuv of constant maQ1iiLu along the circular path, the Oidv chanüc:: his velocity is due to the change in direction of motion. me motion of tne athl moving along a circular path is, therefore, exam Vle of an accelerated motion. Wc. know that the circumference of a cir of radius ris given by 21tr . If ihe athlete t seconds to once around the circular of radius r, the speed v is given by 21tr (8.

##### Scene 14 (45m 55s)

[Audio] MOTION 111 If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This shows that the direction of motion changed at every point when the stone was moving along the circular path. When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above. There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on. Activity _____________ 8.11 • Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in Fig. 8.10. Fig. 8.10: A stone describing a circular path with a velocity of constant magnitude. • Now, let the stone go by releasing the thread. • Can you tell the direction in which the stone moves after it is released? • By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not. What you have learnt • Motion is a change of position; it can be described in terms of the distance moved or the displacement. • The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. • The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time. • The acceleration of an object is the change in velocity per unit time. • Uniform and non-uniform motions of objects can be shown through graphs. • The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely v = u + at s = ut + ½ at2 2as = v2 – u2 2020- 21

MOTION 111 If you carefully note, on being released the stone moves along a straight line tangential to the circular path. This is because once the stone is released, it continues to move along the direction it has been moving at that instant. This shows that the direction of motion changed at every point when the stone was moving along the circular path. When an athlete throws a hammer or a discus in a sports meet, he/she holds the hammer or the discus in his/her hand and gives it a circular motion by rotating his/her own body. Once released in the desired direction, the hammer or discus moves in the direction in which it was moving at the time it was released, just like the piece of stone in the activity described above. There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at constant speed and so on. Activity _____________8.11 • Take a piece of thread and tie a small piece of stone at one of its ends. Move the stone to describe a circular path with constant speed by holding the thread at the other end, as shown in Fig. 8.10. Fig. 8.10: A stone describing a circular path with a velocity of constant magnitude. • Now, let the stone go by releasing the thread. • Can you tell the direction in which the stone moves after it is released? • By repeating the activity for a few times and releasing the stone at different positions of the circular path, check whether the direction in which the stone moves remains the same or not. What you have learnt • Motion is a change of position; it can be described in terms of the distance moved or the displacement. • The motion of an object could be uniform or non-uniform depending on whether its velocity is constant or changing. • The speed of an object is the distance covered per unit time, and velocity is the displacement per unit time. • The acceleration of an object is the change in velocity per unit time. • Uniform and non-uniform motions of objects can be shown through graphs. • The motion of an object moving at uniform acceleration can be described with the help of the following equations, namely v = u + at s = ut + ½ at2 2as = v2 – u2 2020-21

##### Scene 15 (48m 37s)

[Audio] SCIENCE 112 where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity and s is the distance it travelled in time t. • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. Exercises 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C? 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h– 1. What is the average speed for Abdul's trip? 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? 5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? 6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11 2020- 21

SCIENCE 112 where u is initial velocity of the object, which moves with uniform acceleration a for time t, v is its final velocity and s is the distance it travelled in time t. • If an object moves in a circular path with uniform speed, its motion is called uniform circular motion. Exercises 1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? 2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? 3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip? 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? 5. A driver of a car travelling at 52 km h–1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h–1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? 6. Fig 8.11 shows the distance-time graph of three objects A,B and C. Study the graph and answer the following questions: Fig. 8.11 2020-21

##### Scene 16 (51m 0s)

[Audio] MOTION 113 (a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s- 2, with what velocity will it strike the ground? After what time will it strike the ground? 8. The speed-time graph for a car is shown is Fig. 8.12. Fig. 8.12 (a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction. 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth. 2020- 21

MOTION 113

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? 7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s-2, with what velocity will it strike the ground? After what time will it strike the ground? 8. The speed-time graph for a car is shown is Fig. 8.12.

Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? 9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction. 10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

2020-21

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C? A ball is gently dropped from a of 20 m. If its velocity increases urc.r.ormly at the rate 01 10 m so, with what velocity it strike tne ground? Alter what tlrne Wlil It strike tne 'Ihe speed-time graph for a car is shown is Fig. 8.12. 8. 9. 10 Monon 8- g 4 o 2 4 6 (a) Find how far hes car ir (he first 4 seennds. Shade Lhe +he erauh th. •t I epresenLs the dL5L_ance travelled -oy •ar dur1LIC Lhe period. (b) Which pav ot the graph represents uniform motion of State. of the fob Wii g situations are possible and give (a) an object wit a acceleration but with zero velocity (bj an object mov.ng with an acceleration but with uniform (c) moving in certain direction with an ace at-ion in the perpendicular direction. qrtfffcfal satellite is moving in a circular orbit of radius km. Calculate its speed if it takes 24 hours to revolve around the ear

8 g 4. 4 6 8 10 9. 10. Snarle the ar, Ohe szraph th •t epresents the dis travened -oy •ar dunne the period. (b) Which pav or the graph represents uniform motl State of the g situations are possible axu (a) an ubiect a acceleration but with zero ve (b) an Obiect mov.ng vnth an acceleration but with (c) moving in certain direction wit: ace ation in the perpendicular direction. artifieial satellite is moving tn a circular orbit of J km. Catemuate its speed if it takes 24 hours to re around Lhe earth.